Return (terminate) caller function from callee - php

Question

I would like to create a forward-function that terminates the caller-function.
Currently I'm using "exit" in the callee-function to achieve this.
function caller (){
if($condition){
forward();
}
//code that is not executet if $condition is true
}
//current solution
function forward(){
//do something
exit;
}
I would like to omit the return-statement after calling the forward-function.
I guess there is no other way except throwing an exception?
Thanks for your help.

Answer 1

There are several ways:
1) Enclosing the code after if info else branch
2) As you mentioned, throwing an exception
3) or plainly doing:
function caller (){
if($condition){
forward();
return; // you return nothing (void, in PHP's case that would be NULL, I believe)
}
//code that is not executet if $condition is true
}

Answer 2

Throwing exceptions, using exit or some other nasty stuff is overkill no matter how you look at it (if all you want to do is conditional execution of some lines of code).
Solution 1
Use an else statement. This function will implicitly return NULL (in PHP, by default, ommiting the return statement when exiting a function, will return NULL to the caller).
function caller (){
if($condition){
forward();
}
else
{
//code that is not executet if $condition is true
}
}
Solution 2
The return statement will exit the current function.
function caller (){
if($condition){
forward();
return; //this function exits, and no other lines inside this function are executed
}
//code that is not executet if $condition is true
}
Don't use exit to terminate scripts, without passing and exit code. Exit is just like die, except instead of outputting to php://stdout (same thing as what echo does) it will return an exit code to the OS (exit is usefull only in CLI (command line) mode).

Answer 3

You can't terminate a "caller" function before the "called" function terminate itself.
There is a stack of calls that is allocated and return pointer issues.
I suppose that the best way to terminate a function itself is .... just terminate itself.
So
function caller (){
if($condition){
return 0;
}
//code that is not executet if $condition is true
}

Answer 4

I've only seen this answer in the comments, but not in any of the other answers.
To wrap up, let's consider two approaches and why they're not suitable for a forward functionality:
Use exceptions; exceptions should not be used for flow control, they're meant to handle exceptional events (when an operation has failed).
Use goto; this statement can only be used to jump within the same block of execution, which in this case would be inside the caller and the only jump target is at the end of the function.
Use exit(); this will terminate the script altogether and doesn't give any of the underlying code a chance to do anything else. It's typically used for fatal errors of some sort, or in rare cases whereby you set a few HTTP headers and then prevent any more output from being sent to the browser.
Technically speaking, a forward call should basically be used to express "say whatever he says"; and there's a perfect solution for that:
return forward();
It delivers the result of the forward() directly to the code that called the caller() while still respecting the stack execution order of your application.

Answer 5

Possibly, "goto" is the only solution:
function post_to_host($url, $data, $cook, $ref, &$resp_head, $type=1)
{$GLOBALS['POST_TO_HOST.TIME_START']=time();
...
stream_set_timeout($fp, 20);
fputs($fp, "Accept: */*\r\n"); if (check_timeout()) {goto timeout;}
fputs($fp, "Accept-Language: *\r\n"); if (check_timeout()) {goto timeout;}
fputs($fp, "Accept-Encoding: *\r\n"); if (check_timeout()) {goto timeout;}
while (!feof($fp))
{$line.=fgets($fp); if (check_timeout()) {goto timeout;}
}
return $line;
timeout:
{return 'POST EXCEPTION: Total timeout!';
}
}
function check_timeout()
{return time()-$GLOBALS['POST_TO_HOST.TIME_START']>60;
}